int frac{1}{sqrt{2x-x^2}} dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Step $1$: Complete the square for the quadratic expression $2x-x^2$.$2x-x^2 = -(x^2-2x) = -(x^2-2x+1-1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$.Step $2$

Vedclass · Accepted Answer

$\sin^{-1}(x-1) + c$

Vedclass · Answer

(A) Step $1$: Complete the square for the quadratic expression $2x-x^2$.$2x-x^2 = -(x^2-2x) = -(x^2-2x+1-1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$.Step $2$: Rewrite the integral using the completed square.$\int \frac{1}{\sqrt{1-(x-1)^2}} dx$.Step $3$: Use the standard integral formula $\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c$.Let $u = x-1$, then $du = dx$.Therefore, $\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c = \sin^{-1}(x-1) + c$.

$\int \frac{1}{\sqrt{2x-x^2}} dx = $

Similar Questions

Evaluate the integral $\int x^2 \sqrt{8-x^6} \, dx$. Given $|x| < \sqrt{2}$.

If $f(x) = \int \frac{16x^7 + 5x^{10}}{(x^3 + 2 + 3x^8)^2} dx$ for $x \geq 0$ and $f(0) = 1$, then the value of $f(1)$ is

$\int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}=\frac{(\ln (x))^K}{K}+C$
$\Rightarrow 2 K=$

$\int \cos \sqrt{x} \, dx =$ (where $C$ is a constant of integration.)

To find the value of $\int \frac{dx}{x\sqrt{2ax - x^2}}$,the suitable substitution is

Vedclass Test Series

Exam Paper Generator

Online Exam Module